Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, q;
int s, a, b;
vector<pair<double, double>> segs;
double f[N];
double query(int x, int y) {
  double l0 = a + (double)(x - a) * (-s) / (y - s);
  double r0 = b + (double)(x - b) * (-s) / (y - s);
  int p;
  p = upper_bound(segs.begin(), segs.end(), make_pair(l0, (double)0)) -
      segs.begin() - 1;
  double sumL = 0;
  if (p >= 0) {
    sumL += min(segs[p].second, l0) - segs[p].first;
    if (p) sumL += f[p - 1];
  }
  p = upper_bound(segs.begin(), segs.end(), make_pair(r0, (double)0)) -
      segs.begin() - 1;
  double sumR = 0;
  if (p >= 0) {
    sumR += min(segs[p].second, r0) - segs[p].first;
    if (p) sumR += f[p - 1];
  }
  double ret = (sumR - sumL) / (r0 - l0) * (b - a);
  ret = abs(ret);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(7);
  cin >> s >> a >> b;
  cin >> n;
  segs.resize(n);
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    segs[i] = {l, r};
  }
  for (int i = 0; i < n; ++i) {
    f[i] = segs[i].second - segs[i].first;
    if (i) f[i] += f[i - 1];
  }
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    cout << query(x, y) << '\n';
  }
}